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What is the probability density function (PDF) of $f(x)=1/\sin(x)$ when $x$ is uniformly distributed in $(0,90)$?

$f(x)=\sin(x)$ has a known PDF, which has the form $2(\pi\sqrt{1-\sin(x)^2})^{-1}$, but I cannot find the PDF for $1/\sin(x)$. The latter would have interesting applications in astronomy, specially for so-called "luminosity functions".

Thank you very much

Sebastian

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1 Answer 1

up vote 4 down vote accepted

Let $z=1/\sin(x)$, then $x=\sin^{-1}(1/z)$ on $(0,\frac{\pi}{2})$ interval. The pdf you seek, is the differential of uniform c.d.f., which is $\frac{2}{\pi} dx$

$$ \frac{2}{\pi} d( \sin^{-1}( z^{-1} ) ) = \frac{2}{\pi z} \frac{1}{\sqrt{z^2-1}},$$ where variable $z>1$.

By the way, you could have used Mathematica to directly find this:

In[100]:= FullSimplify[PDF[TransformedDistribution[1/Sin[x], 
       x \[Distributed] UniformDistribution[{0, Pi/2}]], z]]

Out[100]= Piecewise[{{2/(Pi*z*Sqrt[-1 + z^2]), z > 1}}, 0]
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Super. Thank you very much, Sasha. –  Sebastian Aug 4 '11 at 16:03

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